Verifying an answer to a question in predicate logic

Question

So, I'm solving a test exam, which doesn't have the answers and this is the question:

Consider the domain of people. Some people are prophets, who may have followers.

$Px$ means that $x$ is a prophet.

$Fxy$ means that $x$ is a follower of $y$.

The predicate $=$ may also be used.

Now write each of the following sentences as a formula of predicate logic:

My answers are the following:

a. All prophets have at least one follower.

a. $∀x~( Px \to ∃y~(¬(x=y) ∧ Fyx) )$

b. There is a prophet who has at least two followers.

b. $∃x~( Px \to ∃y~∃z~(¬(y=z) ∧ Fyx ∧ Fzx) )$

c. There is a prophet who has precisely two followers.

c. $∃x~( Px \to ∃y~∃z~(¬(y=z) ∧ Fyx ∧ Fzx) ∧ ∀t~(Ftx \to ( (t=y) ∨ (t=z) ) ) )$

d. A follower of a prophet also follows any prophet that the first prophet follows.

d. $∀x~∀y~( (Px ∧ Fyx) \to ∀z~( (Pz ∧ Fxz) ∧ Fyz ) )$

Are those even remotely correct?

Answer 1

Your answer to (a) looks fine to me.

For (b) and (c): very close, but you made the same mistakes. First of all, the way you set them up does not guarantee that such prophets exist. If there's at least one non-prophet $x$ among all people (which seems very likely!), then both of your logical statements (b) and (c) are true: for this non-prophet $x$, "$Px$" is false, and therefore any implication of the form "$Px\to\text{whatever}$" is true. So it doesn't say what you want to say. To claim that such a person actually exists, you want to replace implication with conjunction. Second, you forgot to include the condition that followers must be distinct from the prophet, as you did in (a).

For (d), again you have the right idea, but the last part (that follows $\forall z$) isn't correct. You need some implications there to say that if $z$ is a prophet and $x$ follows $z$, then the followers of $x$ follow $z$ — as you can see it's an "if-then" statement in English, so it is a logical implication.

Answer 2

Are those even remotely correct?

Just a few corrections needed.

a. All prophets have at least one follower.

a. $∀x~( Px \to ∃y~(¬(x=y) ∧ Fyx) )$

No need for the inequality, which says the follower is not the prophet. That wasn't specified.

$\forall x~(Px\to\exists y~Fyx)$

Or in Prenex form: $\forall x~\exists y~(Px\to Fyx)$

b. There is a prophet who has at least two followers.

b. $∃x~( Px \to ∃y~∃z~(¬(y=z) ∧ Fyx ∧ Fzx) )$

Not quite.   In this case the inequality specifies that there are two distinct followers, as required.

However, Universal quantifiers have their domain restricted by conditionals, but Existentitals are restricted by conjunction.   You want to say "There is someone who is a prophet and...."

So that is: $∃x~( Px \land ∃y~∃z~(¬(y=z) \land Fyx \land Fzx) )$

or in prenex form: $∃x~∃y~∃z~( Px \land ¬(y=z) \land Fyx \land Fzx)$

c. There is a prophet who has precisely two followers.

c. $∃x~( Px \to ∃y~∃z~(¬(y=z) ∧ Fyx ∧ Fzx) ∧ ∀t~(Fty \to ( (t=y) ∨ (t=z) ) ) )$

Again you have the wrong restriction.   You also need to encompass the $\forall t$ within the scope of the existential variables to which it uses.   And the $t$ should be a follower of $x$.

So: $∃x~( Px \land ∃y~∃z~(¬(y=z) \land Fyx \land Fzx \land ∀t~(Ftx \to ( (t=y) \lor (t=z) ) ) ) )$

or in prenex form: $∃x~∃y~∃z~∀t~ ( Px \land ¬(y=z) \land Fyx \land Fzx \land (Ftx \to ( (t=y) \lor (t=z) ) ) )$

d. A follower of a prophet also follows any prophet that the first prophet follows.

d. $∀x~∀y~( (Px ∧ Fyx) \to ∀z~( (Pz ∧ Fxz) ∧ Fyz ) )$

Close, save for the last universal, which I suspect is a typo.   You wanted to say: for all x,y if y follows x and x is a prophet then for any z if z prophet and x follows z, then y follows z.

$∀x~∀y~( (Px \land Fyx) \to ∀z~( (Pz \land Fxz) \to Fyz ) ) )$

or in prenex form: $∀x~∀y~∀z~( (Px \land Fyx \land Pz \land Fxz) \to Fyz )$